I am given events $A_1, A_2, A_3, \ldots$ such that every two distinct $i,j \in \{1,2,...\}$ satisfy $P(A_i\cap A_j) \leq P(A_i)P(A_j)$.
Why is it that $\sum_{i=1}^n\sum_{j=1}^n P(A_i \cap A_j) \leq \sum_{i,j \in \{1,...,n\}, i \neq j} P(A_i) P(A_j) + \sum_{k=1}^{n} P(A_k)$
Can someone also explain what the sum $\sum_{i,j \in \{1,...,n\}, i \neq j} P(A_i) P(A_j) $ means? Can I write it as a double sum somehow?
The sum in question is over all pairs of indices $(i,j)$ such that $i\neq j$. You could write it as a double sum via: $\sum_{j=1}^n\sum_{i=1,i\neq j}^n$. One other option is to symmetrize it by requiring $j>i$ and write it as $2\sum_{j=2}^{n-1}\sum_{i=1}^{j-1}P(A_i,A_j)$, where we used the fact that $P(A_i,A_j)=P(A_j,A_i)$
So for your question, break $\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)$ into two pieces: $\sum_{j=1}^n\sum_{i=1,i\neq j}^nP(A_i\cap A_j)$ and $\sum_{i=1}^n P(A_i\cap A_i)$. In other words a sum over all indices $(i,j)$ where $i\neq j$ and another sum over indices $(i,j)$ where $i=j$. That should immediately lead to the required upper bound.